3 Answers
3

The latter is because
$$
\frac{x^k-1}{x-1}=x^{k-1}+x^{k-2}+\cdots+1
$$
Now write out the number as
$$ n=\sum_{k=0}^n\overbrace{\ \ \ \ \ d_k\ \ \ \ \ }^{\text{$2$ digit blocks}}100^k$$
and the sum of the two digit blocks as
$$ s=\sum_{k=0}^nd_k$$
then show $99\mid n-s$.

Therefore, if $n = 0 \text{mod}(11)$ then $ a_0 + a_1 + a_2 + a_3 + \cdots + a_{k-1} + a_k = 0 \text{mod}(11)$ and vice versa (each form is equivalent when taken modulo 11) which is to say that $n$ is divisible by $11$ if and only if the sum of its two digit blocks is divisible by $11$.